Question

A particle A of mas of 3mg is balanced by keeping another object of B of charge +30nC below it at a depth of 5cm. What should be the charge on A in order to balance it against gravity?

Answer 1

Given:

A particle A of mas of 3mg is balanced by keeping another object of B of charge +30nC below it at a depth of 5cm.

To find:

Charge on A in order to balance the object against Gravity.

Calculation:

Object A will be balanced against gravity only when the gravitational force acting on it is equal to the electrostatic force exerted by object B.

$\therefore \: F_{g} = F_{E}$

$= > mg = \dfrac{k(q1)(q2)}{ {r}^{2} }$

$= > 3 \times {10}^{ - 6} \times 10 = \dfrac{k(q1)(30 \times {10}^{ - 9} )}{ {(0.05)}^{2} }$

$= > 3 \times {10}^{ - 5} = \dfrac{k(q1)(30 \times {10}^{ - 9} )}{ {(0.05)}^{2} }$

$= > 3 \times {10}^{ - 5} = \dfrac{k(q1)(30 \times {10}^{ - 9} )}{ 25 \times {10}^{ - 4} }$

$= > 3 \times {10}^{ - 5} = \dfrac{k(q1)(6 \times {10}^{ - 9} )}{ 5 \times {10}^{ - 4} }$

$= > {10}^{ - 5} = \dfrac{k(q1)(2 \times {10}^{ - 9} )}{ 5 \times {10}^{ - 4} }$

$= > {10}^{ - 9} = \dfrac{k(q1)(2 \times {10}^{ - 9} )}{ 5 }$

$= > 1 = \dfrac{k(q1)(2 )}{ 5 }$

$= > 1 = \dfrac{9 \times {10}^{9} (q1)(2 )}{ 5 }$

$= > \: q1 = \dfrac{5 \times {10}^{ - 9} }{18}$

$= > \: q1 = 0.27 \times {10}^{ - 9}$

$= > \: q1 = 0.27 \: \: nC$

So, final answer:

$\boxed{ \sf{ \: q1 = 0.27 \: \: nC }}$